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There are 8 blue and 4 white balls in a bag . A ball is drawn at random. Without replacing it another ball is drawn. Find the probability that both the balls drawn are blue ?
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- 7/33
- 14/33
- 11/33
- 2/33
Correct Option: B
Total number of balls in the bag = 12
Probability of drawing one blue ball in the first draw P(E1) = 8C1 / 12C1
= 8/12 = 2/3
After the first drawn of a blue ball, now there are 7 blue and 4 white ball in the bag. Total number of the balls in the bag is 11.
Probability of drawing one blue ball in the second drawn = P(E2) = 7C1 /11C1 = 7/11
∴ Probability that both are blue P(E1 ∩ E2) = P(E1) x P(E2)
= 2/3 x 7/11 = 14 / 33
